What is the best way to disable implicit conversion from pointer types to bool when constructing an std::variant?

Question

Consider the following:

struct foo {
};

struct bar {
};

int main()
{
    foo f;
    bar b;
    std::variant<foo*, bool> v;
    v = &b; // compiles in Visual Studio 19 v16.7.3
}

As discussed in comments, I believe the above is legal C++17. There is a proposal, P0608R3, that was accepted into the standard addressing this kind of surprising behavior, but it was accepted in 2018 (at the San Diego meeting) and thus applies to C++20 not C++17. Further P0608R3 is not currently implemented in Visual Studio, even when compiling to the C++20 preview.

What is the best / least verbose way to make creation of this variant from a pointer that points to a non-foo a compile time error? I believe the following works but is a lot of boilerplate if the variant contains several items.

struct foo {
};

struct bar {
};

using variant_type = std::variant<foo*, bool>;
struct var_wrapper : public variant_type
{
    var_wrapper(foo* v = nullptr) : variant_type(v)
    {}

    var_wrapper(bool v) : variant_type(v)
    {}

    template<typename T>
    var_wrapper(T*) = delete;
};

int main()
{
    foo f;
    bar b;

    var_wrapper vw;
    vw = &f; // fine
    vw = true; // fine
    vw = &b; // compile time error
}

Am I missing some simpler way?

Answer 1

I think the cleanest way to handle implicit conversion doing surprising things with respect to variants is to use a "strong variant" type if the behavior of std::variant is a problem; i.e., implement a variant type that enforces construction only using types that are exactly the types in the variant.

It is easy to test if a type is a member of a parameter pack in C++17 using std::disjunction, leading to an implementation as below:

template<typename... Ts>
class strong_variant : public std::variant<Ts...> 
{
public:
    template <typename T, typename U = 
        typename std::enable_if<std::disjunction_v<std::is_same<T, Ts>...>>::type>
    strong_variant(T v) : std::variant<Ts...>(v)
    {}

    strong_variant() : std::variant<Ts...>()
    {}
};

struct foo {};
struct bar {};

int main()
{
    foo f;
    bar b;
    const foo c_f;

    strong_variant<foo*, std::string, bool> sv;

    sv = &f; // okay.
    sv = true; // okay.
    sv = "foo"s; // okay.

    sv = "foo"; //no, must a string.
    sv = &b;  // no, must be a foo.
    sv = &c_f; // no, must be non-const.
}

Answer 2

Another solution is to introduce another bool wrapper that doesn't construct from anything except from bool:

#include <variant>

struct foo {};
struct bar {};

struct StrongBool {
    bool value = false;

    StrongBool() noexcept = default;
    StrongBool(bool b) noexcept : value(b) {}

    template<class T>
    StrongBool(T) = delete;
};

int main() {
    foo f;
    bar b;
    std::variant<foo*, StrongBool> v;
    v = true;
    v = &f;
    v = &b; // fails to compile
} 

Regardless, limiting acceptable initializers requires introducing type wrappers with user-defined constructors.